1. ## polynomial equation

calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(P(x)-x)^2$

2. Originally Posted by jacks
calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(p(x)-x)^2$
If you put x=0 in this equation then you get $(p(-1))^2+1=0$. This cannot happen if the polynomial has real coefficients. So there are no such real polynomials. If complex coefficients are allowed, then $p(-1)=\pm\sqrt{-1}$, but I don't see where to go from there.

3. Originally Posted by jacks
calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(P(x)-x)^2$
If we presume that p(x) is going to have at least real coefficients we can expland the definig equation to get
$\displaystyle p(x+1) = \frac{1}{2}(x + 1) - \frac{p^2(x)}{2(x + 1)}$

The above says that p(x) must contain a factor of x + 1 and since no real p(0) exists (ala Opalg) we must also require that p(x) contains a factor of x as well, so we have
$\displaystyle p(x) = x(x + 1)q(x)$

where q(x) is a polynomial. This expression needs to be run through the original equation to get restrictions on q(x), but I can't make any sense out of the resulting equation.

-Dan